In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 Ω resistance, and is at a length of 3 m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell is

Question

In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 Ω resistance, and is at a length of 3 m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell is (A) 1.5 Ω (B) 10 Ω (C) 15 Ω (D) 1 Ω

Tardigrade Admin · Accepted Answer

r=((l1-l2/l2)) × R'=((l1-2/2)) × 5 ......(i) and r=((l1-3/3)) × 10 ........(ii) On solving (i) and (ii), r=10 Ω

$1.5 Ω$

$10 Ω$

$15 Ω$

$1 Ω$

$r = (\frac{l _{1} - l _{2}}{l _{2}}) \times R^{'} = (\frac{l _{1} - 2}{2}) \times 5$ ......(i)
and $r = (\frac{l _{1} - 3}{3}) \times 10$ ........(ii)
On solving (i) and (ii), $r = 10 Ω$

Q. In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2m when the cell is shunted by a 5Ω resistance, and is at a length of 3m when the cell is shunted by a 10Ω resistance. The internal resistance of the cell is

1.5Ω

10Ω

15Ω

1Ω

r=(l2​l1​−l2​​)×R′=(2l1​−2​)×5 ......(i) and r=(3l1​−3​)×10 ........(ii) On solving (i) and (ii), r=10Ω

$1.5 Ω$

$10 Ω$

$15 Ω$

$1 Ω$

$r = (\frac{l _{1} - l _{2}}{l _{2}}) \times R^{'} = (\frac{l _{1} - 2}{2}) \times 5$ ......(i)
and $r = (\frac{l _{1} - 3}{3}) \times 10$ ........(ii)
On solving (i) and (ii), $r = 10 Ω$